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Let $K$ be a local field, $K^{sep}$ its separable closure, $G = Gal(K^{sep}/ K)$ the Galois group and $C := \overline{K^{sep}}$ the completion with respect to the induced valuation.

In his paper on $p$-divisible groups, Tate proves that if $K$ is a $p$-adic field, then the continuous cohomology groups $H^{i}(G, C)$ are one-dimensional over $K$ when $i = 0, 1$ and vanish otherwise.

Are these continuous cohomology groups known when $K \simeq \mathbb{F}_{q}((t))$ is non-Archimedean of equal characteristic?

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    $\begingroup$ A local function field? If not, what completion do you want to take? $\endgroup$
    – Will Sawin
    Commented Mar 16, 2021 at 3:38
  • $\begingroup$ Yes, I meant a field of formal Laurent series with the $t$-adic topology, and the completion is the $t$-adic one. $\endgroup$ Commented Mar 16, 2021 at 4:31

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Yes, they are known: They vanish in degrees $i>0$, and for $i=0$ one gets the completed perfection of $K$.

Indeed, let $K'$ be the completed perfection of $K$. Then $G_K=G_{K'}$ as both perfection and completion do not change the etale site. But now $K'$ is already perfectoid, so the same techniques of almost mathematics that Tate uses as an intermediate step in his computation apply to prove that $H^i(G_{K'},C)=0$ for $i>0$, and $H^0(G_{K'},C)=K'$.

[Edit: I realize that it may be worth mentioning that after completion, $\overline{K^{\mathrm{sep}}}$ is perfect.]

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  • $\begingroup$ Thank you, this is very helpful. How do we identify the completed perfection of $K$ with a subspace of invariants in $C$? $\endgroup$ Commented Mar 16, 2021 at 15:36
  • $\begingroup$ As $\overline{K^{\mathrm{sep}}}$ is complete and perfect, there is a map $K'\to C$ (and $C=\overline{K'^{\mathrm{sep}}}$). The Galois cohomology computation is a very special case of Proposition 7.13 here (which shows the stronger statement that $H^i(G_{K'},\mathcal O_C)$ is almost zero for $i>0$, and equal to $\mathcal O_{K'}$ for $i=0$). $\endgroup$ Commented Mar 16, 2021 at 15:40

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